Solve the Differential equation (x^2 + y^2) dx + (x^2 - xy) dy = 0

Answer:[tex]\frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C[/tex]Step-by-step explanation:Given differential equation,[tex](x^2 + y^2) dx + (x^2 - xy) dy = 0[/tex][tex]\implies \frac{dy}{dx}=-\frac{x^2 + y^2}{x^2 - xy}----(1)[/tex]Let y = vxDifferentiating with respect to x,[tex]\frac{dy}{dx}=v+x\frac{dv}{dx}[/tex]From equation (1),[tex]v+x\frac{dv}{dx}=-\frac{x^2 + (vx)^2}{x^2 - x(vx)}[/tex][tex]v+x\frac{dv}{dx}=-\frac{x^2 + v^2x^2}{x^2 - vx^2}[/tex][tex]v+x\frac{dv}{dx}=-\frac{1 + v^2}{1 - v}[/tex][tex]v+x\frac{dv}{dx}=\frac{1 + v^2}{v-1}[/tex][tex]x\frac{dv}{dx}=\frac{1 + v^2}{v-1}-v[/tex][tex]x\frac{dv}{dx}=\frac{1 + v^2-v^2+v}{v-1}[/tex][tex]x\frac{dv}{dx}=\frac{v+1}{v-1}[/tex][tex]\frac{v-1}{v+1}dv=\frac{1}{x}dx[/tex]Integrating both sides,[tex]\int{\frac{v-1}{v+1}}dv=\int{\frac{1}{x}}dx[/tex][tex]\int{\frac{v-1+1-1}{v+1}}dv=lnx + C[/tex][tex]\int{1-\frac{2}{v+1}}dv=lnx + C[/tex][tex]v-2ln(v+1)=lnx+C[/tex]Now, y = vx ⇒ v = y/x[tex]\implies \frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C[/tex]